Question

the time of flight for an object thrown from the rooftop of a tall building is 6.2 seconds. in terms of the initial launch velocity v, and the angle of launch θ, what equation gives you the maximum displacement of the object?
a x=vcos(θ)3.1
b x=vcos(θ)6.2
c x=vsin(θ)3.1
d x=vsin(θ)6.2

Explanation:

Step1: Recall projectile motion formula

In projectile motion, horizontal displacement \( x \) is given by \( x = v_{x} \times t \), where \( v_{x} \) is horizontal component of velocity and \( t \) is time of flight.

Step2: Determine horizontal velocity

Horizontal component of initial velocity \( v \) at angle \( \theta \) is \( v\cos(\theta) \), so \( v_{x}=v\cos(\theta) \).

Step3: Substitute time of flight

Given time of flight \( t = 6.2 \) seconds. Substitute \( v_{x} \) and \( t \) into horizontal displacement formula: \( x=v\cos(\theta)\times6.2 \)? Wait, no, wait the options have 3.1 or 6.2. Wait, maybe the time to reach maximum height? Wait, maximum displacement (horizontal? Or vertical? Wait, the question says "maximum displacement" – but in projectile, horizontal displacement is \( v\cos\theta \times t \), but if it's maximum horizontal displacement? Wait, no, maybe the time given is total time, but maximum horizontal displacement would be when? Wait, no, maybe the question is about horizontal displacement, and the time is total time. Wait the options: A: \( x = v\cos(\theta)3.1 \), B: \( x = v\cos(\theta)6.2 \), C: \( x = v\sin(\theta)3.1 \), D: \( x = v\sin(\theta)6.2 \). Wait, maybe the time to reach maximum height is half of total time? Wait, if total time of flight is 6.2 s, then time to reach max height is \( 6.2/2 = 3.1 \) s. But horizontal velocity is constant, so horizontal displacement at any time is \( v\cos\theta \times t \). But if the question is about maximum horizontal displacement? No, horizontal displacement increases with time. Wait, maybe the question is about vertical displacement? No, vertical displacement formula is different. Wait, maybe the question has a typo, but looking at options, horizontal component is \( v\cos\theta \), and if the time is 3.1 (half of 6.2), maybe the maximum horizontal displacement? No, that doesn't make sense. Wait, maybe the question is about the horizontal range, but in projectile from a height, the range is \( v\cos\theta \times t_{total} \). But the options have 3.1 and 6.2. Wait, maybe the time given is 6.2, but the formula for horizontal displacement (range) is \( v\cos\theta \times t \). But the options: A is \( v\cos\theta \times 3.1 \), B is \( v\cos\theta \times 6.2 \). Wait, maybe the time of flight is 6.2, so total time, so horizontal displacement is \( v\cos\theta \times 6.2 \), but that's option B. Wait, but maybe the question is about maximum height? No, maximum height formula is \( \frac{(v\sin\theta)^2}{2g} \), not matching options. Wait, maybe the question is miswritten, but among options, horizontal component is \( v\cos\theta \), and if the time is 3.1 (half of 6.2, time to reach max height), but horizontal displacement at max height is \( v\cos\theta \times 3.1 \), which is option A. Wait, maybe the question is asking for the horizontal displacement at maximum height (when vertical velocity is zero), which occurs at \( t = t_{total}/2 = 6.2/2 = 3.1 \) s. So horizontal displacement at that time is \( v\cos\theta \times 3.1 \), so option A.

Answer:

A. \( x = v\cos(\theta)3.1 \)